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2 years ago

14

Two cats are mated. One of the parent cats is long haired(recessive allele). The litter which results contains two short haired

and three long haired kittens. What does the second parent look like and what is its genotype?

1 answer:

dexar [7]2 years ago

6 0

Mm I don't know, why is that question? You must have an weird teacher

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Ralph is 3 times as old as Sara. In 6 years, Ralph will be only twice as old as Sara will be then. Find Ralph's age now. Ralph's

Marizza181 [45]

The correct answer is 18.

5 0

2 years ago

Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$

6 0

2 years ago

Solve for e. (-7) = e/3 + 14

raketka [301]

<span>(-7) = e/3 + 14

Subtract 14 on both sides

-7 - 14 = e/3 + 14 - 14

-21 = e/3

Multiply by 3 on both sides

-21 * 3 = e/3 * 3

-63 = e</span>

5 0

2 years ago

A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p

andreyandreev [35.5K]

Answer:

B) 0.894

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that $\mu = 0.55$

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577$

$P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894$

5 0

2 years ago

Can someone please give me the answer

Serga [27]

Answer:

#1 = 234

:)

Step-by-step explanation:

7 0

2 years ago